#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 39 maths Textbook Solution.

Answer:$4 x y z$

Hint Use determinant formula

Given: $\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|=4 x y z$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right| \\ &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 2(y+z) & z & y \\ 2(x+z) & z+x & x \\ 2(x+y) & x & x+y \end{array}\right| \end{aligned}

\begin{aligned} &2 \text { common from } \mathrm{C}_{1}\\ &=2\left|\begin{array}{ccc} (y+z) & z & y \\ (x+z) & z+x & x \\ (x+y) & x & x+y \end{array}\right|\\ &\begin{aligned} &\text { Now } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \\ &=2\left|\begin{array}{ccc} y & z & y \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| \end{aligned} \end{aligned}

\begin{aligned} &\text { Now } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=2\left|\begin{array}{ccc} y & z & 0 \\ 0 & z+x & x \\ y & x & x \end{array}\right| \\ &{ }^{\text {Now }} \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \\ &=2\left|\begin{array}{ccc} y & z & 0 \\ 0 & z+x & x \\ 0 & x-z & x \end{array}\right| \end{aligned}

Expanding w.r.t $C_{1}$

\begin{aligned} &=2 y(z+x) x-x(x-z)-0+0 \\ &=2 y\left(z x+x^{2}-x^{2}+x z\right) \\ &=4 x y z \\ &=R \cdot H . S \end{aligned}